Determine whether the series converges conditionally, absolutely, or not at all.
Look at the series whose terms are the absolute values.
Since |cos n| is between zero and 1,
By the p-test we know that
converges, and by the comparison test we then know that
converges. The original series converges absolutely.
The ratio test looks at the absolute value of the ratio between terms:
This is the same thing as the ratio between the absolute values of the terms:
This means when we use the ratio test, we're really checking the convergence of the series
If the ratio test says "yes, the series converges" it actually means "yes, the original series converges absolutely."
For this particular series, we look at
Since the limit is less than 1, the series converges. Since we were really looking at the terms of , the original series converges absolutely.
This is the classic example of conditional convergence, so we should pay close attention to this answer. We know that converges by the AST,
and we know by the integral test that
does not. This means the series converges conditionally.
Write a real proof that absolute convergence implies convergence.
Assume that converges.
(a) Show that converges.
(b) Show that converges. (hint: an ≤ |an|)
(c) Show that converges.
(a) There's really nothing to do here. Since converges and multiplication by a constant doesn't affect convergence,
converges too. We could also say that since converges, so does
(b) Since ,
Since can't be less than zero (it could be equal, but not less than),
This is just what we need for the comparison test. We know from (a) that converges, so by the comparison test
(c) We know that and converge. This means we can take the difference of the series and get a new series that also converges:
This is what we wanted to show, so we're done.
Make it rain.